673. Number of Longest Increasing Subsequence【Medium】 一维动归
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题目:给一个序列nums,求该序列有多少条最长子序列
思路:最长子序列长度用动归的办法求出:序列对每一个位置i,dp[i]存的是序列从0到i的最长子序列的长度。
状态转移方程: dp[i] = max{dp[k] + 1, 0 <= k < i and nums[k] < nums[i] }
设最长的子序列长度为d
求最长子序列有多少条的办法就是:对每一个dp[i] = d 的位置(即某个最长子序列的末尾),求出以其作为最长子序列的末尾的序列个数
求法为:在求最长子序列的dp过程中,对每个位置i,在求出其从0到i的序列最大长度的同时,求出从所有序列头到i的边数(即序列个数数),用edge[i]来存(这里我解释得不是很清楚,然而我不能更好地解释了,很难把一个道理说得完全清楚)
最后,对所有dp[i] = d 的位置,把edge[i]累积起来就是最终的答案
有一点难
class Solution {public: int findNumberOfLIS(vector<int>& nums) { int l = nums.size(), max = 0; vector<int> dp(l), edge(l); for (int i = 0; i < l; ++i) { dp[i] = 1; edge[i] = 1; for (int j = 0; j < i; ++j) { if (nums[j] < nums[i] && dp[i] < dp[j] + 1) { dp[i] = dp[j] + 1; edge[i] = edge[j];} else if (nums[j] < nums[i] && dp[i] == dp[j] + 1) {edge[i] += edge[j];}}if (max < dp[i]) max = dp[i];}int edges = 0;for (int i = 0; i < l; ++i)if (max == dp[i]) {edges += edge[i];}return edges; }};
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